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4 years ago

Assume that the lines that appear to be tangent are tangent p is the center if each circle find x.

Mathematics

1 answer:

Elodia [21]4 years ago

4 0

Answer:

Step-by-step explanation:

Problem One

All quadrilaterals have angles that add up to 360 degrees.

Tangents touch the circle in such a way that the radius and the tangent form a right angle at the point of contact.

Solution

x + 115 + 90 + 90 = 360

x + 295 = 360

x + 295 - 295 = 360 - 295

x = 65

Problem Two

From the previous problem, you know that where the 6 and 8 meet is a right angle.

Therefore you can use a^2 + b^2 = c^2

a = 6

b =8

c = ?

6^2 + 8^2 = c^2

c^2 = 36 + 64

c^2 = 100

sqrt(c^2) = sqrt(100)

c = 10

x = 10

Problem 3

No guarantees on this one. I'm not sure how the diagram is set up. I take the 4 to be the length from the bottom of the line marked 10 to the intersect point of the tangent with the circle.

That means that the measurement left is 10 - 4 = 6

x and 6 are both tangents from the upper point of the line marked 10.

Therefore x = 6

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3 years ago

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Plot the points on a graph.A(0,0);B(0,2);C(2,2);D(2,0). Join AB, BC,CD and AD. What is the figure formed? Find the area of the f

tia_tia [17]

$\large \green{ \underline{ \blue{ \boxed{\bf{ \red{Given -}}}}}}$

Four points which lie at (0,0) ; (0,2) ; (2,2) ; (2,0)

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The given points lie on graph as per the order given in the attachment.

Which shape?

As the points lie 2 units apart from their adjacent sides. It can be a square or rhombus.

The opposite pairs of sides are also parallel to each other. The adjacent sides of the figure formed are also perpendicular to each other.

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What is the area?

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2 years ago

Please explain your answer. THX!!!!

Sidana [21]

Answer:

1) $cos(x)sin^{2}x$

2)0

Step-by-step explanation:

for part 1:

Given cos(x) - cos^3(x)

= cos(x) (1- cos^2(x))

Using the identity $cos^{2}x +sin^{2}x=1\\ sin^{2}x=1-cos^{2}x$

= cos(x)sin^2(x)

for part 2:

given

= $cos\frac{5\pi }{8}cos\frac{\pi }{8} + sin\frac{5\pi }{8}sin\frac{\pi }{8}$

Using the identity cos(a)cos(b) + sin(a)sin(b)= cos(a-b)

= $cos(\frac{5\pi }{8} -\frac{\pi }{8})$

= $cos\frac{4\pi }{8}$

= $cos\frac{\pi }{2}$

=0!

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3 years ago

A sign 4 meters tall cast a shadow 5 meters long . At the same time, a tree casts a shadow 30 meters long. What is the height of

AfilCa [17]

The information's given in the question are of vital importance. These information's should be very useful while getting to the desired answer of the given question.Let us now concentrate on the problem that needs to be solved.
The height of the sign that casts a shadow of 4 meters is = 5 meters
Then
The height of the tree that casts a shadow of 30 meters = (5/4) * 30 meters
= 150/4 meters
= 37.5 meters
So the tree casting a shadow of 30 meters is actually 37.5 meters in height. I hope the procedure is simple enough for you to understand.

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3 years ago

A building has n floors numbered 1,2,...,n, plus a ground floor g. at the ground floor, m people get on the elevator together, a

fomenos

Let $X_i$ be the random variable indicating whether the elevator does not stop at floor $i$ , with

$X_i=\begin{cases}1&\text{if the elevator does not stop at floor }i\\0&\text{otherwise}\end{cases}$

Let $Y$ be the random variable representing the number of floors at which the elevator does not stop. Then

$Y=X_1+X_2+\cdots+X_{n-1}+X_n$

We want to find $\mathrm{Var}(Y)$ . By definition,

$\mathrm{Var}(Y)=\mathbb E[(Y-\mathbb E[Y])^2]=\mathbb E[Y^2]-\mathbb E[Y]^2$

As stated in the question, there is a $\dfrac1n$ probability that any one person will get off at floor $n$ (here, $n$ refers to any of the $n$ total floors, not just the top floor). Then the probability that a person will not get off at floor $n$ is $1-\dfrac1n$ . There are $m$ people in the elevator, so the probability that not a single one gets off at floor $n$ is $\left(1-\dfrac1n\right)^m$ .

So,

$\mathbb P(X_i=x)\begin{cases}\left(1-\dfrac1n\right)^m&\text{for }x=1\\\\1-\left(1-\dfrac1n\right)^m&\text{for }x=0\end{cases}$

which means

$\mathbb E[Y]=\mathbb E\left[\displaystyle\sum_{i=1}^nX_i\right]=\displaystyle\sum_{i=1}^n\mathbb E[X_i]=\sum_{i=1}^n\left(1\cdot\left(1-\dfrac1n\right)^m+0\cdot\left(1-\left(1-\dfrac1n\right)^m\right)$
$\implies\mathbb E[Y]=n\left(1-\dfrac1n\right)^m$

and

$\mathbb E[Y^2]=\mathbb E\left[\left(\displaystyle\sum_{i=1}^n{X_i}\right)^2\right]=\mathbb E\left[\displaystyle\sum_{i=1}^n{X_i}^2+2\sum_{1\le i$

Computing $\mathbb E[{X_i}^2]$ is trivial since it's the same as $\mathbb E[X_i]$ . (Do you see why?)

Next, we want to find the expected value of the following random variable, when $i\neq j$ :

$X_iX_j=\begin{cases}1&\text{if }X_i=1\text{ and }X_j=1\\0&\text{otherwise}\end{cases}$

If $X_iX_j=0$ , we don't care; when we compute $\mathbb E[X_iX_j]$ , the contributing terms will vanish. We only want to see what happens when both floors are not visited.

$\mathbb P(X_iX_j=1)=\left(1-\dfrac2n\right)^m$
$\implies\mathbb E[X_iX_j]=\left(1-\dfrac2n\right)^m$
$\implies2\displaystyle\sum_{1\le i$

where we multiply by $n(n-1)$ because that's how many ways there are of choosing indices $i,j$ for $X_iX_j$ such that $1\le i$ .

So,

$\mathrm{Var}[Y]=n\left(1-\dfrac1n\right)^m+2n(n-1)\left(1-\dfrac2n\right)^m-n^2\left(1-\dfrac1n\right)^{2m}$

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3 years ago

Assume that the lines that appear to be tangent are tangent p is the center if each circle find x. ​

Assume that the lines that appear to be tangent are tangent p is the center if each circle find x.